To determine if it is possible to transform the initial configuration of piles into 105 piles with one stone each, let's first examine the operations allowed and their effects on the total number of piles:

1. Combining Two Piles: When any two piles are combined into one pile, the total number of piles decreases by one.

2. Dividing a Pile: When a pile consisting of an even number of stones is divided into two equal piles, the total number of piles increases by one.

Initially, we have 3 piles with sizes 51, 49, and 5 stones. The sum of stones is:
$$51 + 49 + 5 = 105.$$
To achieve the goal of having 105 piles with one stone each, we need to ensure that each pile contains exactly one stone.

Step-by-step Analysis:

- Begin with 3 piles.

- Objective: Increase the number of piles to 105.

- Key Insight: Notice that in order to end up with 105 piles of one stone each, we need each stone to be in its own separate pile.

Using the allowed operations to achieve this:

- Dividing piles:
- Dividing a pile results in 2 piles, and since the sum of stones remains constant, we aim to transform existing stones into individual smaller piles.
- Ideally, each division would result in making a new additional pile when dividing even (or evenly splittable) numbers, thus increasing total pile count.

- Initially, we have 3 piles. To turn this into 105 piles, the total number needs to be increased by:
$$105 - 3 = 102 \text{ additional piles}.$$

Since each division of initially available piles consisting of even-numbered stones increases the number of piles by one, we need 102 such division operations. This means we need to start with sufficient even pile divisible configurations.

- Combining piles only decreases the pile count and hence does not help in moving from 3 to 105 piles.

- Feasibility Explanation:
- We have piles starting at odd counts (51, 49, 5), limiting our division activities at the outset.
- If any number of these operations don't exhaustively lead through intermediate even numbers towards the number 105 as piles, this would be unsolvable.
- It's impossible to arrange such steps from initial uneven piles given - any subsequent numbers fulfilling divisibility fail without alternating combing to some extent.

Hence, given combining cannot exceed sum reductions effectively and is counter to goal, piling to achieve 105 isn't possible.

Thus, from our understanding and confirmation by operation exploration, the answer is:
$$\boxed{\text{No}}$$